Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
import java.util.Scanner;

public class Main {
  public static void main(String[] args) {
    Scanner scanner = new Scanner(System.in);

    int N = scanner.nextInt();
    double[] p = new double[N + 1];
    for (int i = 1; i <= N; i++) p[i] = scanner.nextDouble();

    double[][] dp = new double[N + 1][N + 1];
    dp[0][0] = 1.d;
    for (int i = 1; i <= N; i++) {
      for (int j = 0; j <= i; j++) {
        dp[i][j] = dp[i - 1][j] * (1 - p[i]);
        if (j > 0) dp[i][j] += dp[i - 1][j - 1] * p[i];
      }
    }
    double res = 0.d;
    for (int i = N / 2 + 1; i <= N; i++) res += dp[N][i];
    System.out.println(res);
  }
}